9514 1404 393
Answer:
x = {-0.5-√1.25, -0.5+√1.25, 2, -0.5+i√0.75, -0.5-i√0.75}
Step-by-step explanation:
I like to use a graphing calculator to find clues as to the roots of higher-degree polynomials. Here, we see that x=2 is the only real rational root. Dividing that out by synthetic division, we see the remaining quartic factor is ...
2x^5 -6x^3 -4x^2 -2x +4 = 0
2(x -2)(x^4 +2x^3 +x^2 -1) = 0
We can recognize that the quartic factor is actually the difference of two squares:
x^4 +2x^3 +x^2 -1 = (x^2 +x)^2 -1 = 0
So it resolves to two quadratic factors.
(x^2 +x +1)(x^2 +x -1) = 0
One will have real roots, as shown by the graph. The other will have complex roots.
x^2 +x + 1/4 = 1 +1/4 . . . . complete the square for the factor with real roots
(x +1/2)^2 = 5/4
x = -1/2 ± √(5/4) . . . . . . irrational real roots
__
x^2 +x = -1 . . . . . . . . . . the quadratic factor with complex roots
(x +1/2)^2 = -1 +1/4 . . . complete the square
x = -1/2 ± i√(3/4) . . . . irrational complex roots
__
In summary, the values of x that satisfy the equation are ...
x = 2
x = -1/2 ± √(5/4)
x = -1/2 ± i√(3/4)
